\newproblem{lay:4_3_32}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.3.32}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $V$ and $W$ be vector spaces, and $T:V\rightarrow W$ a linear transformation between the two. Let $S=\{\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_p\}$ be a subset of $V$.
	Suppose $T$ is a one-to-one transformation, so that an equation $T(\mathbf{u})=T(\mathbf{v})$ always implies $\mathbf{u}=\mathbf{v}$. Show that if the set of images
	$T(S)=\{T(\mathbf{v}_1),T(\mathbf{v}_2),...,T(\mathbf{v}_p)\}$ is linearly dependent, then $S=\{\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_p\}$ is also linearly dependent. This fact
	shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly
	dependent).
}{
  % Solution
	If $T(S)$ is linearly dependent, then there exist coefficients $c_1$, $c_2$, ..., $c_p$ not all of them zero such that
	\begin{center}
		$c_1T(\mathbf{v}_1)+c_2T(\mathbf{v}_2)+...+c_pT(\mathbf{v}_p)=\mathbf{0}_W$
	\end{center}
	If $T$ is one-to-one, we infer that it must also be
	\begin{center}
		$c_1\mathbf{v}_1+c_2\mathbf{v}_2+...+c_p\mathbf{v}_p=\mathbf{0}_V$
	\end{center}
	that is, there exist coefficients $c_1$, $c_2$, ..., $c_p$ not all of them zero such that the linear combination of the transformed vectors is $\mathbf{0}_V$. This
	means that $S$ is a set of linearly dependent vectors.
}
\useproblem{lay:4_3_32}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
